(3e)x=xy+x2+y2 Differentiate both parts with respect to x . Use the Chain Rule
dxd((3e)x)=dxd(xy+x2+y2)
(ln3+1)(3e)x=y+xdxdy+2x+2ydxdy
Solve for dxdy
dxdy=x+2y(ln3+1)(3e)x−2x−y
Find the slope of the tangent line at point (0,1)
slope=m=dxdy∣(0,1)
=0+2(1)(ln3+1)(3e)0−2(0)−1=2ln3
The equation of the tangent line in point-slope form
y−1=2ln3(x−0) The equation of the tangent line in slope-intercept form
y=2ln3⋅x+1
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